Calculating the equilibrium constant K

AP Chemistry· AP Chemistry CED — Equilibrium· 14 min read

1. Calculating K from Known Equilibrium Amounts★★☆☆☆⏱ 4 min

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The simplest K calculation uses directly given equilibrium concentrations or partial pressures. By the law of mass action, for a general balanced reaction $a A + b B ⇌ c C + d D$ , the equilibrium constant expression follows reaction stoichiometry.

K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} \quad \text{or} \quad K_p = \frac{(P_C)^c (P_D)^d}{(P_A)^a (P_B)^b}

📐 Worked Example

A 2.0 L reaction vessel contains an equilibrium mixture for the reaction $N_{2} (g) + O_{2} (g) ⇌ 2 N O (g)$ . At equilibrium, the mixture has 0.12 mol $N_{2}$ , 0.18 mol $O_{2}$ , and 0.12 mol $N O$ . Calculate $K_{c}$ for this reaction.

1
Convert mole values to equilibrium molar concentrations by dividing moles by reaction volume:
$[N_2] = \frac{0.12\ \text{mol}}{2.0\ \text{L}} = 0.060\ M, \quad [O_2] = \frac{0.18\ \text{mol}}{2.0\ \text{L}} = 0.090\ M, \quad [NO] = \frac{0.12\ \text{mol}}{2.0\ \text{L}} = 0.060\ M$
2
Write the correct $K_{c}$ expression, confirm no pure solids/liquids are included:
$K_c = \frac{[NO]^2}{[N_2][O_2]}$
3
Substitute values and calculate with correct significant figures:
$K_c = \frac{(0.060)^2}{(0.060)(0.090)} = \frac{0.0036}{0.0054} \approx 0.67$

Exam tip:

Always convert moles to molar concentration before plugging into $K_{c}$ ; MCQ traps often use distractors that match results from raw mole values.

2. Calculating K Using ICE Tables★★★☆☆⏱ 5 min

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Most AP exam problems do not give all equilibrium values directly. Instead, you use an ICE (Initial-Change-Equilibrium) table to map stoichiometric changes and solve for unknown equilibrium values. The ICE table has three core rows:

Initial: Concentrations/pressures before the reaction begins to reach equilibrium
Change: Change in concentration as the system approaches equilibrium, following stoichiometry: negative for reactants consumed, positive for products formed, with $x$ as the unknown change
Equilibrium: Final equilibrium value, calculated as Initial + Change

Once you use the known equilibrium value to solve for $x$ , you can find all other equilibrium values and calculate $K$ .

📐 Worked Example

0.50 mol of $N O C l$ is placed in a 1.0 L reaction vessel and decomposes via $2 N O C l (g) ⇌ 2 N O (g) + C l_{2} (g)$ . At equilibrium, the concentration of $C l_{2}$ is 0.12 M. Calculate $K_{c}$ .

1
Set up the ICE table, using $x$ as the change in concentration of $C l_{2}$ :
2
Species Initial (M) Change (M) Equilibrium (M)
$N O C l$ 0.50 $- 2 x$ $0.50 - 2 x$
$N O$ 0 $+ 2 x$ $2 x$
$C l_{2}$ 0 $+ x$ $x$
3
Use the given equilibrium $[C l_{2}]$ to solve for $x$ :
$[Cl_2]_{eq} = 0.12\ M = x \implies x = 0.12$
4
Calculate all equilibrium concentrations:
$[NOCl]_{eq} = 0.50 - 2(0.12) = 0.26\ M, \quad [NO]_{eq} = 2(0.12) = 0.24\ M, \quad [Cl_2]_{eq} = 0.12\ M$
5
Substitute into the $K_{c}$ expression and calculate:
$K_c = \frac{[NO]^2[Cl_2]}{[NOCl]^2} = \frac{(0.24)^2(0.12)}{(0.26)^2} \approx 0.10$

Species	Initial (M)	Change (M)	Equilibrium (M)
$N O C l$	0.50	$- 2 x$	$0.50 - 2 x$
$N O$	0	$+ 2 x$	$2 x$
$C l_{2}$	0	$+ x$	$x$

Exam tip:

Always match the coefficient of $x$ in the change row to the stoichiometric coefficient of the species. If 2 moles of reactant are consumed, the change is $- 2 x$ , not $- x$ .

3. Relating Kc and Kp, and Manipulating K for Modified Reactions★★★☆☆⏱ 4 min

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Two common exam problems require converting between $K_{c}$ and $K_{p}$ , and finding the new $K$ when a reaction is reversed, scaled, or combined. The relationship between $K_{c}$ and $K_{p}$ comes from the ideal gas law:

K_p = K_c(RT)^{\Delta n}

Where $Δ n = moles of gaseous products - moles of gaseous reactants$ , $R = 0.0821 L \cdot a t m / (m o l \cdot K)$ , and $T$ is absolute temperature in Kelvin. Three core rules apply for modifying K:

Reversing a reaction: $K_{n e w} = \frac{1}{K _{or i g ina l}}$
Multiplying all coefficients by a factor $n$ : $K_{n e w} = (K_{or i g ina l})^{n}$
Adding two reactions to get a total reaction: $K_{t o t a l} = K_{1} \times K_{2}$

📐 Worked Example

Given the reaction $2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$ has $K_{c} = 2.8 \times 1 0^{2}$ at 1000 K. (a) Calculate $K_{p}$ for this reaction. (b) Calculate $K$ for the reaction $S O_{3} (g) ⇌ S O_{2} (g) + \frac{1}{2} O_{2} (g)$ at 1000 K.

1
For part (a), calculate $Δ n$ , only counting gaseous species:
$\Delta n = 2 - (2+1) = -1$
2
Substitute into the conversion formula:
$K_p = K_c(RT)^{\Delta n} = (2.8 \times 10^2)(0.0821 \times 1000)^{-1} = \frac{280}{82.1} \approx 3.4$
3
For part (b), the new reaction is the original reversed and scaled by $\frac{1}{2}$ . First reverse the reaction:
$K = \frac{1}{2.8 \times 10^2} \approx 3.57 \times 10^{-3}$
4
Scale by $\frac{1}{2}$ by raising $K$ to the $\frac{1}{2}$ power:
$K = \sqrt{3.57 \times 10^{-3}} \approx 0.060$

Exam tip:

$Δ n$ only counts gaseous species; do not include aqueous solutes, pure solids, or pure liquids when calculating $Δ n$ for the Kc-Kp conversion.

4. Calculating K from Standard Gibbs Free Energy Change★★★★☆⏱ 3 min

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This is a common cross-unit topic connecting Unit 7 (Equilibrium) and Unit 9 (Thermodynamics). The relationship between standard Gibbs free energy change $Δ G^{\circ}$ and $K$ is:

\Delta G^\circ = -RT \ln K

Rearranging to solve for $K$ gives:

\ln K = -\frac{\Delta G^\circ}{RT}

Key note: $Δ G^{\circ}$ must be in units of J/mol to match $R = 8.314 J / (m o l \cdot K)$ . If $Δ G^{\circ} < 0$ , $ln K$ is positive so $K > 1$ (products favored at equilibrium); if $Δ G^{\circ} > 0$ , $K < 1$ (reactants favored).

📐 Worked Example

At 298 K, the standard Gibbs free energy change for the dissociation of formic acid $H C O O H (a q) ⇌ H C O O^{-} (a q) + H^{+} (a q)$ is $Δ G^{\circ} = + 21.3 k J / m o l$ . Calculate the acid dissociation constant $K_{a}$ at 298 K.

1
Convert $Δ G^{\circ}$ from kJ/mol to J/mol to match units of $R$ :
$21.3\ kJ/mol \times 1000\ J/kJ = 21300\ J/mol$
2
Substitute into the rearranged formula:
$\ln K = -\frac{\Delta G^\circ}{RT} = -\frac{21300}{(8.314)(298)} \approx -8.60$
3
Exponentiate to solve for $K$ :
$K = e^{-8.60} \approx 1.8 \times 10^{-4}$
4
Confirm consistency: a positive $Δ G^{\circ}$ for a weak acid dissociation gives $K < 1$ , which matches the expected behavior of a weak acid.

✓ Quick check

Test your understanding with this AP-style multiple choice question:

For the reaction $2 H g O (s) ⇌ 2 H g (l) + O_{2} (g)$ , the equilibrium partial pressure of $O_{2}$ at 350°C is 0.48 atm. What is $K_{p}$ for this reaction?
- 0.48
- 0.23
- 0.92
- 2.1
Reveal answer
0.48 —
Correct! Pure solids (HgO) and pure liquids (Hg) are omitted from the K expression because their activity is 1, so $K_{p} = P_{O_{2}} = 0.48$ .

Exam tip:

Always convert $Δ G^{\circ}$ from kJ/mol to J/mol before plugging into the formula. Using kJ directly will give a K that is orders of magnitude off, costing points on FRQs.

5. Common Pitfalls

Wrong move:

Including pure solids and pure liquids in the K expression, using their concentrations to calculate K

Why:

Students often plug all species from the balanced equation into the expression, forgetting that activity of pure condensed phases is 1

Correct move:

Cross out any pure solid or pure liquid every time you write a K expression before calculating

Wrong move:

Using raw mole values directly in Kc calculations instead of converting to molar concentration

Why:

Problems often give moles and volume, and students skip the division step because it seems trivial

Correct move:

Explicitly calculate concentration = moles / volume before plugging into Kc

Wrong move:

Calculating Δn as reactant moles minus product moles, or counting non-gaseous species in Δn

Why:

Students mix up the order of subtraction, or forget only gaseous species count for Δn

Correct move:

Explicitly write Δn = moles gaseous products - moles gaseous reactants before using the Kp formula

Wrong move:

When scaling a reaction by a factor of 2, multiplying K by 2 instead of squaring K

Why:

Students confuse scaling reaction coefficients with scaling the equilibrium constant, mixing addition with exponentiation

Correct move:

Remind yourself: if every coefficient is multiplied by n, K is raised to the nth power

Wrong move:

Leaving ΔG° in kJ/mol when calculating K from ΔG°

Why:

ΔG° is almost always reported in kJ/mol, but R uses Joules in the standard formula

Correct move:

Always check units of R and ΔG°, multiply ΔG° by 1000 to convert kJ to J before plugging in

Wrong move:

In ICE tables, writing a change of -x for a species with a stoichiometric coefficient of 2

Why:

Students default to x as the change for all species regardless of stoichiometry

Correct move:

Write the change for each species as (stoichiometric coefficient) × x, with a negative sign for reactants, before solving

6. Quick Reference Cheatsheet

Category	Formula	Notes
$K_{c}$ from equilibrium concentrations	$K_{c} = \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}}$	Omit pure solids/liquids; $[X]$ = equilibrium molar concentration
$K_{p}$ from equilibrium partial pressures	$K_{p} = \frac{( P _{C} ) ^{c} ( P _{D} ) ^{d}}{( P _{A} ) ^{a} ( P _{B} ) ^{b}}$	Only includes gaseous species; $P_{X}$ = equilibrium partial pressure
$K_{c}$ to $K_{p}$ conversion	$K_{p} = K_{c} (R T)^{Δ n}$	$Δ n$ = moles gaseous products - moles gaseous reactants; $R = 0.0821$
$K$ for reversed reaction	$K_{n e w} = \frac{1}{K _{or i g ina l}}$	Reversing the reaction inverts K
$K$ for scaled reaction	$K_{n e w} = (K_{or i g ina l})^{n}$	$n$ = factor all coefficients are multiplied by
$K$ for summed reactions	$K_{t o t a l} = K_{1} \times K_{2}$	Multiply K values when adding reaction steps
$K$ from $Δ G^{\circ}$	$ln K = - \frac{Δ G ^{\circ}}{R T}$	$Δ G^{\circ}$ in J/mol; $R = 8.314$ ; $Δ G^{\circ} < 0 ⟹ K > 1$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Calculate K from ICE table
2022 · FRQ
Convert Kc to Kp, manipulate K
2021 · FRQ
Calculate K from ΔG°

Going deeper

unit overviewUnit 7 Equilibrium Overview

What's Next

Mastering calculation of K is the foundational prerequisite for all subsequent equilibrium topics in AP Chemistry, starting with using the reaction quotient Q to predict the direction a reaction will shift to reach equilibrium, then solving for equilibrium concentrations from initial conditions and a known K. Without the ability to correctly calculate K, you cannot solve problems involving Le Chatelier’s principle, acid-base titrations, buffer solutions, or solubility equilibria, which together make up over 15% of the total AP Chemistry exam score. This topic also connects the thermodynamics concept of Gibbs free energy to equilibrium, which is a common cross-unit FRQ topic on the exam.